Problem 52 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find four numbers in A.P. whose sum is 32 and sum of squares is 276 .

Step-by-Step Solution

Answer

The two sets of four numbers that satisfy the given conditions are \(\frac{9}{2}, \frac{23}{6}, \frac{37}{6}, \frac{51}{6}\) and \(\frac{15}{2}, \frac{31}{6}, \frac{33}{6}, \frac{35}{6}\).

Step 1: Define the given information

First, let's define what we are given: 1. Four numbers are in an arithmetic progression (A.P.) 2. Sum of the numbers is 32 3. Sum of the squares of the numbers is 276 Let the four numbers be \(a, a + d, a + 2d, a + 3d\), where \(a\) is the first term and \(d\) is the common difference.

Step 2: Write the given equations

We can write two equations from the given information. 1. Sum of these four numbers is 32, so \(a + (a + d) + (a + 2d) + (a + 3d) = 32\) 2. Sum of the squares of these numbers is 276, so \(a^2 + (a + d)^2 + (a + 2d)^2 + (a + 3d)^2 = 276\)

Step 3: Simplify the first equation

Simplify the first equation to find a relation between \(a\) and \(d\): \((4a + 6d) = 32\) Divide both sides by 2: \((2a + 3d) = 16\) Now, let's express \(a\) in terms of \(d\): \(a = -\frac{3}{2}d + 8\)

Step 4: Substitute the expression for a in the second equation

Replace \(a\) in the second equation with the expression we just found: \[\left(-\frac{3}{2}d + 8\right)^2 + \left(-\frac{3}{2}d + 8 + d\right)^2 + \left(-\frac{3}{2}d + 8 + 2d\right)^2 + \left(-\frac{3}{2}d + 8 + 3d\right)^2 = 276\]

Step 5: Simplify the equation and solve for d

Simplify the equation and solve for \(d\): \(9d^2 - 48d + 64 = 0\) Divide both sides by 3: \(3d^2 - 16d + 21 = 0\) This is a quadratic equation in \(d\). Using the quadratic formula to solve for \(d\), we get: \(d = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(3)(21)}}{2(3)}\) \(d = \frac{16 \pm \sqrt{256 - 252}}{6}\) \(d = \frac{16 \pm \sqrt{4}}{6}\) Now, we have two possible values for \(d\): \(d = \frac{14}{6} = \frac{7}{3}\) or \(d = \frac{2}{6} = \frac{1}{3}\)

Step 6: Find the corresponding values for a

Using the expression for \(a\) and plugging in the values of \(d\), we can find the corresponding values of \(a\): 1. When \(d = \frac{7}{3}\), \(a = -\frac{3}{2} \times \frac{7}{3} + 8 = -\frac{21}{6} + 8 = \frac{48 - 21}{6} = \frac{27}{6} = \frac{9}{2}\) 2. When \(d = \frac{1}{3}\), \(a = -\frac{3}{2} \times \frac{1}{3} + 8 = -\frac{3}{6} + 8 = \frac{45}{6} = \frac{15}{2}\)

Step 7: Write the four numbers for both cases

Now, we can write the four numbers for both cases: 1. When \(a = \frac{9}{2}\) and \(d = \frac{7}{3}\), the four numbers are: \[\frac{9}{2}, \frac{9}{2} + \frac{7}{3}, \frac{9}{2} + 2 \times \frac{7}{3}, \frac{9}{2} + 3 \times \frac{7}{3}\] \[= \frac{9}{2}, \frac{23}{6}, \frac{37}{6}, \frac{51}{6}\] 2. When \(a = \frac{15}{2}\) and \(d = \frac{1}{3}\), the four numbers are: \[\frac{15}{2}, \frac{15}{2} + \frac{1}{3}, \frac{15}{2} + 2 \times \frac{1}{3}, \frac{15}{2} + 3 \times \frac{1}{3}\] \[= \frac{15}{2}, \frac{31}{6}, \frac{33}{6}, \frac{35}{6}\] Thus, the two sets of four numbers that satisfy the given conditions are \(\frac{9}{2}, \frac{23}{6}, \frac{37}{6}, \frac{51}{6}\) and \(\frac{15}{2}, \frac{31}{6}, \frac{33}{6}, \frac{35}{6}\).

Key Concepts

Sum of Squares

When working with an arithmetic progression (A.P.), we may sometimes need to calculate the sum of the squares of the terms. The sum of squares is a valuable tool, especially in problems involving quadratics. Here, the sum of squares of four numbers in A.P. is provided as 276. This gives us a clue on how we can set up our mathematical expressions and solve the given problem. For numbers in A.P., each term increases by a fixed common difference, and their typical representation involves the first term, commonly denoted as \(a\), and the common difference \(d\).

To calculate the sum of squares of numbers in A.P., write each term: \(a, a + d, a + 2d, a + 3d\), then square and sum them:

• \(a^2\)
• \((a + d)^2\)
• \((a + 2d)^2\)
• \((a + 3d)^2\)

Adding these up gives the equation for the sum of squares: \(a^2 + (a + d)^2 + (a + 2d)^2 + (a + 3d)^2 = 276\). This equation will help us derive a quadratic expression later on.

Quadratic Equation

Quadratic equations are vital in solving problems involving arithmetic progressions, especially when additional constraints like sum of squares are present. A quadratic equation is a polynomial equation of degree 2, typically written as \(ax^2 + bx + c = 0\).

In our problem, when we substitute the expression of \(a\) derived from the sum of the terms equation into the sum of squares equation, it simplifies into a quadratic equation in terms of \(d\):\[3d^2 - 16d + 21 = 0\]

Solving this quadratic equation gives us the possible values for the common difference \(d\). These values of \(d\) indicate the various progressions that meet the original problem’s criteria. Solving quadratic equations can be done using methods such as:

• Factoring, if straightforward factors exist.
• Completing the square to perfect the quadratic form.
• Using the quadratic formula: \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our case, the quadratic formula is used to find \(d = \frac{7}{3}\) or \(d = \frac{1}{3}\), leading to different sets of A.P. numbers.

Solving Equations

Solving equations is a fundamental skill in mathematics, allowing us to find unknown values. In this exercise, we need to solve equations derived from the conditions given - four numbers in A.P. with specific sum and sum of squares.

First, express the condition for the sum of the numbers (\(a + a+d + a+2d + a+3d = 32\)) to find a relationship between \(a\) and \(d\). Simplify this equation:

• Combine like terms: \(4a + 6d = 32\)
• Solve for \(a\) in terms of \(d\): \(a = -\frac{3}{2}d + 8\)

Next, substitute this expression for \(a\) into the second equation concerning the sum of squares to find values for \(d\). From the quadratic formula, we find \(d\), and then substitute back to find \(a\).

Once \(a\) and \(d\) are determined, we calculate the four numbers in A.P. Solving these equations step-by-step helps verify potential values against the problem’s constraints, ensuring we find all solutions that satisfy both the sum and sum of squares.